On Wed, Nov 13, 2002 at 06:13:14PM -0500, Buddha Buck wrote:
> Definition: A "ballot" consists of a ranking A>B>C>D>... of options
> submitted by a voter. It defines a total ordering of options for a
> particular voter (i.e., for any pair of options A and B, we can claim
> that a particular voter feels either A>B, A<B, or A=B, and iff A>B, then
> B<A).
Ok.
Note that A=B only happens for the case where the voter mentioned neither
A nor B in their ballot. In my opinion, we should ignore the number A=B:
it should have no bearing on the result.
> Let |A>B| be the number of voters who voted A>B.
> Similarly, for |A<B| and |A=B|.
>
> (Obviously, |A>B| + |A=B| + |A<B| = total number of voters, because of
> the definition of a ballot.)
Ok.
> > 2. Options which do not defeat the default option are eliminated.
> >
> > Definition: Option A defeats option B if more voters prefer optio
> > A over option B than prefer option B over option A.
>
> Let A>>B ("A defeats B") if |A>B| > |B>A|
> Let A==B ("A ties B") if |A>B| = |B>A|
> Let A<<B ("A is defeated by B") if |A>B| < |B>A|
> (Note: A==A, for all options A)
>
> Eliminate all options A if Default>>A.
>
> Clarification: What if Default==A?
Then A does not defeat the default option.
> > 3. If an option has a quorum requirement, that option must defeat
> > the default option by the number of votes specified in the quorum
> > requirement or the option is eliminated.
>
> Does is mean:
>
> Eliminate A if |A>Default| < Quorum
>
> or
>
> Eliminate A if |A>Default| - |Default>A| < Quorum
It means "Eliminate A if |A>Default| - |Default>A| < Quorum". You
should have been able to determine this the text I proposed by
examining the definition of "defeat".
> Again, how do we deal with that |A=Default| case?
See above.
> >
> > 4. If an option has a supermajority requirement, that option must
> > defeat the default option by the ratio of votes specified in the
> > quorum requirement or the option is eliminated.
>
> (?) Eliminate A if |A>Default| / |A<Default| < Supermajority Ratio
Yes. The original language in the constitution makes that clear, I'll
include some of that language in my next draft.
> Again, what about the |A=Default| votes?
> >
> > 5. If one remaining option defeats any other remaining options,
> > that option wins.
>
> s/any/all/
Oops, yes.
> "Condorcet Winner"
>
> If there is a remaining option A, such that for all remaining options B,
> either A=B, or A>>B.
A=B is a tie -- ties are recognized in a seperate procedural step.
I suppose I could add some rules to recognize ties in additional
places, but in my opinion that's unnecessary complexity.
> > 6. If more than one option remains after the above steps, we use
> > Cloneproof Schultz Sequential Dropping to eliminate any cyclic
> > ambiguities and then pick the winner. These represent a procedure
> > and must be carried out in the specified order:
> >
> > i. All options not in the Schultz set are eliminated.
> >
> > Definition: An option C is in the Schultz set if there is no
> > other option D such that C is in the beat path of D AND D is
> > not in the beat path of C.
>
> Let A>>>B mean there is a possibly empty sequence C, D, ..., E, F of
> remaining options such that A>>C, C>>D, ..., E>>F, F>>B
>
> Then B is on the beat path of A.
>
> The Schultz Set = { A | A>>>A }
No, because the schultz set includes ties.
> Note: Because A==A, it isn't the case that A>>A, so if the Schultz set
> includes A, then there must be a B!=A such that A>>B>>...>>A. Since
> A>>B, we then have B>>...>>A>>B, so B is also in the Schultz set.
> Therefore, the Schultz Set can't be a Singleton Set.
I don't agree with this reasoning. However, once we have a singleton set
we have a winner, so we could almost treat your conclusion as true.
> > Definition: An option F is in the beat path of option G if
> > option G defeats option F or if there is some other option
> > H where option H is in the beat path of G AND option F is in
> > the beat path of H.
> >
> > ii. Unless this would eliminate all options in the Schultz set,
> > the options which have the weakest defeat are eliminated.
> >
> > Definition: The strength of a defeat is represented by two
> > numbers: the number of votes for the defeated option and the
> > number of votes for the defeating option.
> >
> > The more votes in favor of a defeated option, the weaker
> > the defeat. Where two pairs of options have the same number
> > of votes in favor of the defeated option, the fewer votes in
> > favor of the defeating option, the weaker the defeat.
>
> So if we have two defeats A>>B and C>>D, then we lexigraphically compare
> (|B>A|, -|A>B|) and (|D>C|, -|C>D|)
>
> What do you mean by "options with the weakest defeat"?
I meant |A>B| and |B>A| where (|B>A|, -|A>B|) is a weakest defeat,
[I didn't phrase that very will in the draft.]
> My understanding was that we were removing defeats from consideration.
> If, for example, there were four options A, B, C, D in the Schultz Set,
> we'd initially be looking at the following set of defeats, in strongest
> to weakest order:
>
> A>>B
> A>>D
> B>>C
> C>>A
> D>>C
> B>>D
>
> After eliminating the weakest defeat (in this case, B>>D, we no longer
> consider it when determining the Schultz Set, as if we had declared
> B==D, so that neither B>>D or D>>B held.
In essence, yes. More specifically, |B>D|=|D>B|=0.
> So what do we really want to eliminate here?
We want to eliminate that option pair from consideration when determining
the next Schwartz set.
> And...
>
> What if we had |D>C| = |B>D|, |D=C| = |B=D|, |D<C| = |B<D|, so that
> neither D>>C nor B>>D was weaker than the other? Do we eliminate both
> defeats?
Yes -- this is very important if there's only a small number of voters
participating. Also, I used this in how I phrased the rule to recognize
ties.
> > iii. If eliminating the weakest defeat would eliminate all options
> > in the Schultz set, a tie exists and the person with the
> > casting vote picks from among these options.
>
> Hmmm, if we had:
>
> Defeats = {A>>B, B>>C, C>>A}
>
> then the Schultz set is {A, B, C}. Eliminating the weakest defeat
> (C>>A) would break all the cycles, so !(A>>>A), !(B>>>B), !(C>>>C), so
> the Schultz set is now empty. Do we want to the casting-vote-caster to
> decide in this case, or do we want to say that A wins?
The schultz set is not empty. A is in the schultz set, and A wins.
> I think it would be better to declare a tie iff all the DEFEATS would be
> eliminated, not all the options. Since all the defeats would be
> eliminated only if they were all of equal strength, there shouln't be a
> problem.
A tie exists after all defeats have been eliminated, not before.
> > iv. Otherwise, a new schultz set is found, with those weakest
> > defeats eliminated,
> >
> > v. If this new schultz set contains only one option, that option
> > wins.
>
> As defined, the Schultz set can't be singleton. Maybe I'm
> misinterpreting teh Schultz set.
Yes.
Thanks,
--
Raul
